# Pricing of $(S(T_0)-S(T))^+$

Problem: Consider a new derivative that at time $$T$$ pays $$Y =(S(T_0) − S(T))^+$$

where $$0 < T_0 < T$$ is a fixed date.

(i) Show that the arbitrage-free of Y at time $$t = T_0$$ is given by $$\pi_{T_0} (Y ) =pS(T_0)$$ where p is independent of the stock price.

(ii) Determine the arbitrage-free price of derivative $$Y$$ at time $$t < T_0$$.

To value the contract I would use the Black-Scholes PDE solution $$E^{Q}((S(T_0) − S(T))^+|\mathscr{F}_{T_{0}})=E^{Q}((S(T_0) − S(T)) \:1_{S(T_0\geqslant S(T))}|\mathscr{F}_{T_{0}})$$

I know that $$S(T_0)-S(T)$$ is independent of $$\mathscr{F}_{T_0}$$. But it is not independent of $$1_{S(T_0\geqslant S(T))}$$, so I cannot split them into two expected values. Even if I did, I would not get the result. I tried to solve using the exponential form of S(T) but got nowhere.

I though that when $$t=T_0$$ then $$Y$$ is a put option which implies that $$E^{Q}((S(T_0) − S(T))^+|\mathscr{F}_{T_{0}})=S(T_0)(e^{T-T_0}\Phi(-d_2)-\Phi(-d_1))$$ where d_2 and d_1 do not depend on the stock.

Question:

Is my solution correct?

How do I solve the second question when $$t? I think I cannot used a pre-defined result.

At time $$T_0$$, the strike price becomes known and the option turns into a normal'' put option, i.e. \begin{align*} V(T_0,S_{T_0}) &= S_{T_0}e^{-r(T-T_0)}\Phi(-d_2)-S_{T_0}e^{-q(T-T_0)}\Phi(-d_1) \\ &= S_{T_0}\underbrace{\left(e^{-r(T-T_0)}\Phi(-d_2)-e^{-q(T-T_0)}\Phi(-d_1)\right),}_{=:p} \end{align*} where $$p$$ is indeed independent of the stock price because \begin{align*} d_{1,2}=\frac{r-q\pm\frac{1}{2}\sigma^2}{\sigma}\sqrt{T-T_0}. \end{align*}
In general, for $$t> T_0$$, the Black-Scholes formula is \begin{align*} V(t,S_t) &= S_{T_0}e^{-r(T-t)}\Phi(-d_2)-S_te^{-q(T-t)}\Phi(-d_1), \end{align*} where \begin{align*} d_{1,2}=\frac{\ln\left(\frac{S_t}{S_{T_0}}\right)+\left(r-q\pm\frac{1}{2}\sigma^2\right)(T-t)}{\sigma\sqrt{T-t}}. \end{align*} Of course, $$\lim\limits_{t\downarrow T_0}V(t,S_t)=S_{T_0}p$$.
For $$t, the option value is \begin{align*} V(t,S_t)&=e^{-r(T-t)}\mathbb{E}^\mathbb{Q}_t\left[\max\{S_{T_0}-S_T,0\}\right] \\ &=e^{-r(T-t)}\mathbb{E}^\mathbb{Q}_t\left[\mathbb{E}^\mathbb{Q}_{T_0}\left[\max\{S_{T_0}-S_T,0\}\right]\right] \\ &=e^{-r(T_0-t)}\mathbb{E}^\mathbb{Q}_t\left[e^{-r(T-T_0)}\mathbb{E}^\mathbb{Q}_{T_0}\left[\max\{S_{T_0}-S_T,0\}\right]\right]\\ &=e^{-r(T_0-t)}\mathbb{E}^\mathbb{Q}_t\left[S_{T_0}p\right] \\ &=pe^{-r(T_0-t)}S_te^{(r-q)(T_0-t)} \\ &=S_te^{-q(T_0-t)}\left(e^{-r(T-T_0)}\Phi(-d_2)-e^{-q(T-T_0)}\Phi(-d_1)\right). \end{align*} Of course, $$\lim\limits_{t\uparrow T_0}V(t,S_t)=S_{T_0}p$$.
• As @StackG correctly adds, the option is pricable in closed form for many models with closed-form solution for European put options. To keep the calculations simple, we'd need $p$ to be independent of $S_{T_0}$. This applies to exponential Lévy processes (e.g. Merton and Kou's jump diffusion models as well as NIG, VG and CGMY) and stochastic volatility models (e.g. Heston or Grasselli's 4/2 model). – Kevin Feb 20 at 23:36